How do you use substitution to integrate $\frac{x}{1 - x}$ $x/(1-x)$ ?

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Answer 1 · 2018-04-18T20:41:13

$\int (\frac{x}{1 - x}) d x$ $int ( x/(1-x) ) dx$

$u = 1 - x$ $u = 1- x$

$d u = - d x$ $du = -dx$

$d x = - d u$ $dx = -du$

Substitute back in

$\int (\frac{x}{u}) \cdot - d u$ $int ( x/u )* -du$

We still have an $x$ $x$ in the problem, so let's use our $u$ $u$ substitution to solve for $x$ $x$ :

$u = 1 - x$ $u = 1-x$

$x = 1 - u$ $x = 1-u$

Now we have

$- \int (\frac{1 - u}{u}) \cdot d u$ $-int ( (1-u)/u )* du$

$- \int (\frac{1}{u} - \frac{u}{u})$ $- int ( 1/u - u/u )$

Split it up

$- \int (\frac{1}{u}) \cdot d u + \int (\frac{u}{u}) d u$ $- int ( (1)/u )* du + int (u/u) du$

$- \int (\frac{1}{u}) \cdot d u + \int d u$ $- int ( (1)/u )* du + int du$

$- ln | u | + u$ $- ln|u| + u$

Get back in terms of $x$ $x$

$- ln | 1 - x | + 1 - x$ $-ln | 1-x | + 1- x$

How do you use substitution to integrate x1−xx/(1-x)?